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Math Help - Another Surface Area question

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    Another Surface Area question

    we have z=1+3x+2y^2 and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
    So far:
    dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA=  \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA, so the integral looks ugly...
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    Quote Originally Posted by apw235@nyu.edu View Post
    we have z=1+3x+2y^2 and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
    So far:
    dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA=  \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA, so the integral looks ugly...
    Have you sketched the region of integration in the xy-plane?

    Your integral is S = \int_{x = 0}^{x = 2} \int_{y = x/2}^{y = 1} \sqrt{16y^2+10} \, dy \, dx.

    Life is much easier if you reverse the order of integration:

    S = \int_{y = 0}^{y = 1} \int_{x = 0}^{x = 2y} \sqrt{16y^2+10} \, dx \, dy.

    This should cause little trouble.
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